Consider the following equation.
$$7y-3=-3(4-x)$$
Step 2 of 2 : Find the equation of the line which passes through the point (6, - 2) and is parallel to the given line. Express your answer in slope-intercept form. Simplify your answer:

Step 1 ：Consider the following equation: $7y-3=-3(4-x)$.

Step 2 ：First, we need to rearrange the equation to the form $y=mx+c$, where $m$ is the slope of the line.

Step 3 ：Doing so, we get $y=\frac{3}{7}x-\frac{9}{7}$.

Step 4 ：From this, we can see that the slope of the line is $\frac{3}{7}$.

Step 5 ：We are asked to find the equation of the line which passes through the point (6, -2) and is parallel to the given line.

Step 6 ：Since parallel lines have the same slope, the slope of the new line will also be $\frac{3}{7}$.

Step 7 ：We can use the point-slope form of the line equation, $y-y1=m(x-x1)$, where $(x1,y1)$ is the point through which the line passes, to find the equation of the line.

Step 8 ：Substituting the given point (6, -2) and the slope $\frac{3}{7}$ into the equation, we get $y=\frac{3}{7}x-\frac{32}{7}$.

Step 9 ：Final Answer: The equation of the line which passes through the point (6, - 2) and is parallel to the given line is $\boxed{{\displaystyle y=\frac{3}{7}x-\frac{32}{7}}}$.